I'm trying to find out all $z \in C$ that satisfy the following condition:

$|z+1|+|z-i|=3$

I understand that $|z|=r$ represents a circle with a radius of $r$.
I also understand that $|z+1|=r$ can be written as $|(x+1)+yi|=\sqrt{(x+1)^2+y^2}=r$ which can then be squared to get $(x+1)^2+y^2=r^2$ which represents the circle with a radius of r, with center in $(-1,0)$.

So, back to my problem:

Unlike the example with $|z+1|=r$, where it is easy to square the equation, squaring
$|z+1|+|z-i|=3$ written as $\sqrt{(x+1)^2+y^2}+\sqrt{x^2+(y-1)^2}=3$ equals:

$(x+1)^2+(y-1)^2+x^2+y^2+2\sqrt{(x+1)^2+y^2}\sqrt{x^2+(y-1)^2}=9$ which is a nightmare to solve, if at all possible.

I am sure there must be some elegant way to solve this kind of problem. The way I'm thinking is this:

$|z+1|$ by itself seems to represent a circle centered at $(-1,0)$, with an undefined radius, and $|z-i|$ seems to represent a circle centered at $(0,1)$, also with an undefined radius. However, the sum of those two radii must be 3. But I can't seem to wrap my mind around what this would represent (when drawn on Cartesian plane), or how to solve it analytically. So, how would one approach this problem? Even a hint would (probably) suffice.

$|z+1|$ is the distance from the point $z$ to the point $-1$, and $|z-i|$ is the distance from $z$ to the point $i$. Thus, you’re looking at the set of all points $z$ such the sum of the distances from $z$ to $-1$ and $i$ is $3$. Which conic section is defined as the locus of points $P$ such that the sum of $|PA|$ and $|PB|$ is a constant for two fixed points $A$ and $B$?